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Multiple modulus inequalities
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19 Dec 2009, 01:01
Hi guys
Need to know how to solve multiple modulus inequalities
Ex  Solve x10  2x+3 <= 5x10
I do understand that we need to first find the roots, which happen to be 10, 3/2 and 2 in this case
we can then divide them into intervals 
x<3/2 3/2<x<2 2<x<10 x>10
what do we do post this division into intervals?



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Re: Multiple modulus inequalities
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19 Dec 2009, 02:47
zaarathelab wrote: Hi guys
Need to know how to solve multiple modulus inequalities
Ex  Solve x10  2x+3 <= 5x10
I do understand that we need to first find the roots, which happen to be 10, 3/2 and 2 in this case
we can then divide them into intervals 
x<3/2 3/2<x<2 2<x<10 x>10
what do we do post this division into intervals? If we dot he way you are proposing: Yes, there are 3 check point at which absolute values in the example change their sign: 3/2, 2, and 10 (te values of x for which the expression in  equals to zero). Then you should expand the modulus in these ranges: A. \(x<\frac{3}{2}\) > \(x102x+3\leq{5x10}\) > \((x10)+(2x+3)\leq{(5x10)}\) > \(6x\leq{3}\) > \(x\leq{\frac{1}{2}\) > as we considering the range when \(x<\frac{3}{2}\), then finally we should write \(x<\frac{3}{2}\): (\(\frac{3}{2}\))B. \(\frac{3}{2}\leq{x}\leq{2}\) > \((x10)(2x+3)\leq{(5x10)}\) > \(2x<=3\) > \(x<={\frac{3}{2}\) > \({\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}\); (\(\frac{3}{2}\))(\(\frac{3}{2}\))C. \(2<x<10\) > \((x10)(2x+3)\leq{5x10}\) > \(17\leq{8x}\) > \(x\geq{\frac{17}{8}}\) > \({\frac{17}{8}}\leq{x}<10\); (\(\frac{17}{8}\))(\(10\))D. \(x\geq{10}\) > \((x10)(2x+3)\leq{5x10}\) > \(3\leq{6x\) > \(x\geq{\frac{1}{2}}\) > \(x\geq{10}\); (\(\frac{17}{8}\))(\(10\))After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are: \(x\leq{\frac{3}{2}}\) and \({\frac{17}{8}}\leq{x}\) (\(\frac{3}{2}\))(\(\frac{3}{2}\))(\(\frac{17}{8}\))(\(10\))One important detail when you define the critical points (10, 3/2 and 2) you should include them in either of intervals you check. So when you wrote: x<3/2 3/2<x<2 2<x<10 x>10 You should put equal sign for each of the critical point in either interval. For example: x =<3/2 3/2<x< =2 2<x<1 =0 x>10 Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality.
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Re: Multiple modulus inequalities
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01 Feb 2012, 01:31
Could somebody explain this in detail, please?
I didn't get the concept of roots here, it doesn't make the equation zero  isn't it?



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Re: Multiple modulus inequalities
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01 Feb 2012, 01:47



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Re: Multiple modulus inequalities
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Updated on: 01 Feb 2012, 02:05
Hi, I solved the modulus by taking the positive & negative scenarios of each modulus. I solved them separately & arrived at 1/2 <= x & x<= 1/2. Am I on the right track or messed up some where??
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Last edited by boomtangboy on 01 Feb 2012, 02:05, edited 1 time in total.



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Re: Multiple modulus inequalities
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01 Feb 2012, 01:59



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Re: Multiple modulus inequalities
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04 Oct 2012, 10:45
Bunuel wrote: zaarathelab wrote: Hi guys
Need to know how to solve multiple modulus inequalities
Ex  Solve x10  2x+3 <= 5x10
I do understand that we need to first find the roots, which happen to be 10, 3/2 and 2 in this case
we can then divide them into intervals 
x<3/2 3/2<x<2 2<x<10 x>10
what do we do post this division into intervals? If we dot he way you are proposing: Yes, there are 3 check point at which absolute values in the example change their sign: 3/2, 2, and 10 (te values of x for which the expression in  equals to zero). Then you should expand the modulus in these ranges: A. \(x<\frac{3}{2}\) > \(x102x+3\leq{5x10}\) > \((x10)+(2x+3)\leq{(5x10)}\) > \(6x\leq{3}\) > \(x\leq{\frac{1}{2}\) > as we considering the range when \(x<\frac{3}{2}\), then finally we should write \(x<\frac{3}{2}\): (\(\frac{3}{2}\))B. \(\frac{3}{2}\leq{x}\leq{2}\) > \((x10)(2x+3)\leq{(5x10)}\) > \(2x<=3\) > \(x<={\frac{3}{2}\) > \({\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}\); (\(\frac{3}{2}\))(\(\frac{3}{2}\))C. \(2<x<10\) > \((x10)(2x+3)\leq{5x10}\) > \(17\leq{8x}\) > \(x\geq{\frac{17}{8}}\) > \({\frac{17}{8}}\leq{x}<10\); (\(\frac{17}{8}\))(\(10\))D. \(x\geq{10}\) > \((x10)(2x+3)\leq{5x10}\) > \(3\leq{6x\) > \(x\geq{\frac{1}{2}}\) > \(x\geq{10}\); (\(\frac{17}{8}\))(\(10\))After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are: \(x\leq{\frac{3}{2}}\) and \({\frac{17}{8}}\leq{x}\) (\(\frac{3}{2}\))(\(\frac{3}{2}\))(\(\frac{17}{8}\))(\(10\))One important detail when you define the critical points (10, 3/2 and 2) you should include them in either of intervals you check. So when you wrote: x<3/2 3/2<x<2 2<x<10 x>10 You should put equal sign for each of the critical point in either interval. For example: x =<3/2 3/2<x< =2 2<x<1 =0 x>10 Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality. Hi Brunel, How do we decide the signs in front of the modulus in the equation. Eg:: x10  2x+3 <= 5x10......say we are looking in an interval  3/2 to 2. then based on what concentration do we put +or  sign in front of the modulus. and please explain >> step 1 x = sqrt (x^2) ? < step w => x = x ? How do this. please explain concept behind this math. Thanks!!
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Re: Multiple modulus inequalities
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05 Oct 2012, 03:50



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Multiple modulus inequalities
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11 Jan 2019, 08:49
Bunuel wrote: zaarathelab wrote: Hi guys
Need to know how to solve multiple modulus inequalities
Ex  Solve x10  2x+3 <= 5x10
I do understand that we need to first find the roots, which happen to be 10, 3/2 and 2 in this case
we can then divide them into intervals 
x<3/2 3/2<x<2 2<x<10 x>10
Bun
what do we do post this division into intervals? If we dot he way you are proposing: Yes, there are 3 check point at which absolute values in the example change their sign: 3/2, 2, and 10 (te values of x for which the expression in  equals to zero). Then you should expand the modulus in these ranges: A. \(x<\frac{3}{2}\) > \(x102x+3\leq{5x10}\) > \((x10)+(2x+3)\leq{(5x10)}\) > \(6x\leq{3}\) > \(x\leq{\frac{1}{2}\) > as we considering the range when \(x<\frac{3}{2}\), then finally we should write \(x<\frac{3}{2}\): (\(\frac{3}{2}\))B. \(\frac{3}{2}\leq{x}\leq{2}\) > \((x10)(2x+3)\leq{(5x10)}\) > \(2x<=3\) > \(x<={\frac{3}{2}\) > \({\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}\); (\(\frac{3}{2}\))(\(\frac{3}{2}\))C. \(2<x<10\) > \((x10)(2x+3)\leq{5x10}\) > \(17\leq{8x}\) > \(x\geq{\frac{17}{8}}\) > \({\frac{17}{8}}\leq{x}<10\); (\(\frac{17}{8}\))(\(10\))D. \(x\geq{10}\) > \((x10)(2x+3)\leq{5x10}\) > \(3\leq{6x\) > \(x\geq{\frac{1}{2}}\) > \(x\geq{10}\); (\(\frac{17}{8}\))(\(10\))After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are: \(x\leq{\frac{3}{2}}\) and \({\frac{17}{8}}\leq{x}\) (\(\frac{3}{2}\))(\(\frac{3}{2}\))(\(\frac{17}{8}\))(\(10\))One important detail when you define the critical points (10, 3/2 and 2) you should include them in either of intervals you check. So when you wrote: x<3/2 3/2<x<2 2<x<10 x>10 You should put equal sign for each of the critical point in either interval. For example: x =<3/2 3/2<x< =2 2<x<1 =0 x>10 Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality. Bunuel, I believe that I follow along entirely until you reach the point where you are determining the viability of the solutions. How have you checked the viability of the solutions? In this tutorial:https://gmatclub.com/forum/mathabsolutevaluemodulus86462.html It seems to teach that one must simply determine if the solution is within the range they are checking and if not, then it is not viable. So, in this example, would x<=1/2 satisfy the criteria of x<3/2 because x would have to be less than 1/2 to be less than 3/2? Also, how do you conclude that the inequality does not hold between 3/2 and 17/8? Again, I was able to arrive at the same numbers, but I am uncertain how you figured that the inequality does not hold in that interval. Sorry as I am a bit lost! Thank you in advance for your help! Best, NCH2024



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Re: Multiple modulus inequalities
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12 Jan 2019, 02:55
nch2024 wrote: Bunuel wrote: zaarathelab wrote: Hi guys
Need to know how to solve multiple modulus inequalities
Ex  Solve x10  2x+3 <= 5x10
I do understand that we need to first find the roots, which happen to be 10, 3/2 and 2 in this case
we can then divide them into intervals 
x<3/2 3/2<x<2 2<x<10 x>10
Bun
what do we do post this division into intervals? If we dot he way you are proposing: Yes, there are 3 check point at which absolute values in the example change their sign: 3/2, 2, and 10 (te values of x for which the expression in  equals to zero). Then you should expand the modulus in these ranges: A. \(x<\frac{3}{2}\) > \(x102x+3\leq{5x10}\) > \((x10)+(2x+3)\leq{(5x10)}\) > \(6x\leq{3}\) > \(x\leq{\frac{1}{2}\) > as we considering the range when \(x<\frac{3}{2}\), then finally we should write \(x<\frac{3}{2}\): (\(\frac{3}{2}\))B. \(\frac{3}{2}\leq{x}\leq{2}\) > \((x10)(2x+3)\leq{(5x10)}\) > \(2x<=3\) > \(x<={\frac{3}{2}\) > \({\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}\); (\(\frac{3}{2}\))(\(\frac{3}{2}\))C. \(2<x<10\) > \((x10)(2x+3)\leq{5x10}\) > \(17\leq{8x}\) > \(x\geq{\frac{17}{8}}\) > \({\frac{17}{8}}\leq{x}<10\); (\(\frac{17}{8}\))(\(10\))D. \(x\geq{10}\) > \((x10)(2x+3)\leq{5x10}\) > \(3\leq{6x\) > \(x\geq{\frac{1}{2}}\) > \(x\geq{10}\); (\(\frac{17}{8}\))(\(10\))After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are: \(x\leq{\frac{3}{2}}\) and \({\frac{17}{8}}\leq{x}\) (\(\frac{3}{2}\))(\(\frac{3}{2}\))(\(\frac{17}{8}\))(\(10\))One important detail when you define the critical points (10, 3/2 and 2) you should include them in either of intervals you check. So when you wrote: x<3/2 3/2<x<2 2<x<10 x>10 You should put equal sign for each of the critical point in either interval. For example: x =<3/2 3/2<x< =2 2<x<1 =0 x>10 Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality. Bunuel, I believe that I follow along entirely until you reach the point where you are determining the viability of the solutions. How have you checked the viability of the solutions? In this tutorial:https://gmatclub.com/forum/mathabsolutevaluemodulus86462.html It seems to teach that one must simply determine if the solution is within the range they are checking and if not, then it is not viable. So, in this example, would x<=1/2 satisfy the criteria of x<3/2 because x would have to be less than 1/2 to be less than 3/2? Also, how do you conclude that the inequality does not hold between 3/2 and 17/8? Again, I was able to arrive at the same numbers, but I am uncertain how you figured that the inequality does not hold in that interval. Sorry as I am a bit lost! Thank you in advance for your help! Best, NCH2024 3/2 < 1/2, so x's which are less than 3/2 are also less than 1/2. (\(\frac{3}{2}\))\((\frac{1}{2})\)
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: Multiple modulus inequalities
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15 Jan 2019, 05:34
Bunuel wrote: Bunuel wrote: Hi guys
Need to know how to solve multiple modulus inequalities
Ex  Solve x10  2x+3 <= 5x10
I do understand that we need to first find the roots, which happen to be 10, 3/2 and 2 in this case
we can then divide them into intervals 
x<3/2 3/2<x<2 2<x<10 x>10
3/2 < 1/2, so x's which are less than 3/2 are also less than 1/2.
(\(\frac{3}{2}\))\((\frac{1}{2})\) [/quote] Thank you, Bunuel!




Re: Multiple modulus inequalities &nbs
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15 Jan 2019, 05:34






